have the same cosine as the arco has: but if, instead of the equation +=2.cos. +=2.cos.( (+0) or =2.cos. (4+0) or = 2cos. (2+0) m m m or =2.cos. (2-0) or =2.cos. (4-0) or =2 cos. ((2n+2)-0) m m the resulting expressions will be respectively =cos. (2π + 0) or =2.cos. (4π + 0) or = 2 cos. (2 η π+0); xm+ xm or = 2.cos. (2 π θ), or = 2 cos. (4 π - θ), or =2.cos. (2n + 2) π – 0, which expressions, by what has just appeared, (see 11. 1, 2, &c.) are divisors; in other words, xem 2 cos.0.x + 1 may be re presented by a product, of which these latter quantities are the factors; accordingly, If we make 0=0, we have cos. 0=1, and cos. π 4π (x-1).(x2-2.COS. 2x + 1)(x - cos. + 1). &c. m m x+1 which is the analytical expression of Cotes's Theorem. See Harmonia Mensurarum, p. 114, &c. and De Moivre's Miscellanea Analytica, p. 17, &c. We will now proceed to investigate expressions for the powers of the cosine and sine of an arc, in terms of the cosines and sines of multiple arcs, which expressions are highly useful in all Mathematical Investigations connected with Physical Astronomy. PROBLEM 7. It is required to express the powers of the cosine and sine of an arc, in terms involving the cosines and sines of the multiple arc. (collecting into pairs the terms equidistant from each extremity of the series) * For the expansion of the Binomial, see Wood's Algebra, page 109, first edition; or Vince's Fluxions, p. 45; or Woodhouse's Principles of Analytical Calculation, page 24, &c. ... by page 48, 2n-1 cos." A = cos. n A + n. cos. (n-2) A+n. n-1 cos. (n - 4) A + &c. The number of terms is n + 1, therefore, if n be even, the consequently, since cos. (n - n) A = 1, the last term x 1.3.5 n=2, 2. cos. A = cos. 2A + 1. n=3, 22.cos. A = cos. 3 A + 3. cos. A n=4, 23.cos.4 A = cos. 4A + 4. cos. 2 A + 3. ..... [e"] [e] ..[e] n=5, 24. cos.5 A = cos. 5 A +5.cos. 3 A + 10 cos. A... [e"] n=6, 25.cos. A = cos. 6 A + 6 cos. 4 A +15.cos. 2A +10 [e] In order to obtain a general form for sin." A, in the form for π cos." A, substitute instead of A, - A, then A π 2-1.cos." (-4) = cos. (n-nA)+n.cos. (n-2) (-4) 2 + n(n-1).cos. (n-4)(-1) + &c. 2 A A Now, let n be even, and of the form 22m.p, p being an odd number, or let n be, as it is called, pariter par; number, and therefore cos. (2m-p. 2 π) is = 1. π 1 Again, (n-2) = (2amp-2) = (2am -p – 1) π, but (22m-p-1) is an odd number, and..cos. (@2m-p-1) = -1; since cos. (pm - nA)=cos.pw.cos. n A = cos. n A, 2-1. sin." A = cos. n A + n. cos. (n - 2) A and, in both these cases, the last term is, as before, = sin. &c. = ± sin. n A, where the upper sign is to be taken, if n be 1, 5, 9, &c. Hence, ±2"-1. sin." A. = n.sin. (n 2) A + n.(n-1) sin. (n-4) A- &c. 1.2 n=5, 24. sin.5 A = sin. 5 A 5. sin. 3A + 10 sin. A.... [f"] n=6, 25.sin.6 A = - cos. 6 A+6 cos. 4A – 15 cos. 2A +10 [f"] PROBLEM 8. It is required to express the tangent of twice, thrice, &c. an arc in terms of the tangent of the simple arc. and, by a similar method, since 4 A = 3A + A, and 5 A = 4 A + A, 1-3.tan. A 1 tan. A |